Question

anyone please fast answer this

Answer 1

this is the correct answer

Answer 2

$\sqrt{ {sec \: b}^{2} + {cosec \: b}^{2} } = \sqrt{1 + {tan \: b}^{2} + 1 + {cot \: b}^{2} } \\ \\ = \sqrt{ {tan \: b}^{2} + {cot}^{2} + 2 \: tan \: b \: \times cot \: b } \\ \\ \sqrt{ {(tan \: b + \: cot \: b)}^{2} } = tan \: b + \: cot \: b$


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